可以使用
std::unique
和通过 std::make_move_iterator
函数创建的迭代器吗?我尝试了以下代码,并且成功了。#include <iostream>
#include <ostream>
#include <vector>
#include <algorithm>
#include <limits>
#include <iterator>
#include <cstdlib>
struct A
{
A() : i(std::numeric_limits< double >::quiet_NaN()) { std::cout << __PRETTY_FUNCTION__ << "\n"; }
A(double ii) : i(ii) { std::cout << __PRETTY_FUNCTION__ << "\n"; }
A(A const & a) : i(a.i) { std::cout << __PRETTY_FUNCTION__ << "\n"; }
A(A && a) : i(std::move(a.i)) { std::cout << __PRETTY_FUNCTION__ << "\n"; a.i = std::numeric_limits< double >::quiet_NaN(); }
A & operator = (A const & a) { std::cout << __PRETTY_FUNCTION__ << "\n"; i = a.i; return *this; }
A & operator = (A && a) { std::cout << __PRETTY_FUNCTION__ << "\n"; i = std::move(a.i); a.i = std::numeric_limits< double >::quiet_NaN(); return *this; }
bool operator < (A const & a) const { std::cout << __PRETTY_FUNCTION__ << "\n"; return (i < a.i); }
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wfloat-equal"
bool operator == (A const & a) const { std::cout << __PRETTY_FUNCTION__ << "\n"; return (i == a.i); }
#pragma clang diagnostic pop
friend
std::ostream &
operator << (std::ostream & o, A const & a)
{
return o << a.i;
}
private :
double i;
};
int
main()
{
std::vector< A > v{1.0, 1.0, 2.0, 2.0, 2.0, 3.0, 4.0, 4.0, 5.0, 6.0, 6.0, 7.0};
std::cout << "v constructed\n\n\n\n";
std::sort(v.begin(), v.end());
auto const end = std::unique(std::make_move_iterator(v.begin()), std::make_move_iterator(v.end())).base();
std::copy(v.begin(), end, std::ostream_iterator< A >(std::cout, " "));
std::cout << std::endl;
return EXIT_SUCCESS;
}
但也许这是与实现相关的成功呢?
另外,还有来自<numeric>
和<algorithm>
的其他算法呢?
std::unique
无论如何都移动,所以我看不出你从中获得了什么好处。 - T.C.std::unique
的复制赋值在此处。 - Tomilov AnatoliyMoveAssignable
类型,因此对于库实现者来说,不移动实际上需要更多的工作。 - T.C.